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NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.4 are part of NCERT Solutions for Class 10 Maths. Here are we have given Chapter 14 Statistics Class 10 NCERT Solutions Ex 14.4.
| Board | CBSE |
| Textbook | NCERT |
| Class | Class 10 |
| Subject | Maths |
| Chapter | Chapter 14 |
| Chapter Name | Statistics |
| Exercise | Ex 14.4 |
| Number of Questions Solved | 3 |
| Category | NCERT Solutions |
NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.4
NCERT Solutions for Class 10 Maths
Question 1.
The following distribution gives the daily income of 50 workers of a factory.
| Daily income (in Rs) | 100 – 120 | 120 – 140 | 140 – 160 | 160 – 180 | 180 – 200 |
| Number of workers | 12 | 14 | 8 | 6 | 10 |
Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.
We can find frequency distribution table of less than type as following –
| Daily income (in Rs)(upper class limits) |
Cumulative frequency |
| Less than 120 | 12 |
| Less than 140 | 12 + 14 = 26 |
| Less than 160 | 26 + 8 = 34 |
| Less than 180 | 34 + 6 = 40 |
| Less than 200 | 40 + 10 = 50 |
Now taking upper class limits of class intervals on x-axis and their respective frequencies on y-axis we can draw its ogive as following –
Question 2.
During the medical check-up of 35 students of a class, their weights were recorded as follows:
| Weight (in kg) | Number of students |
| Less than 38 | 0 |
| Less than 40 | 3 |
| Less than 42 | 5 |
| Less than 44 | 9 |
| Less than 46 | 14 |
| Less than 48 | 28 |
| Less than 50 | 32 |
| Less than 52 | 35 |
Draw a less than type ogive for the given data. Hence obtain the median weight from the graph verify the result by using the formula.
Solution:
The given cumulative frequency distributions of less than type is –
| Weight (in kg) upper class limits |
Number of students (cumulative frequency) |
| Less than 38 | 0 |
| Less than 40 | 3 |
| Less than 42 | 5 |
| Less than 44 | 9 |
| Less than 46 | 14 |
| Less than 48 | 28 |
| Less than 50 | 32 |
| Less than 52 | 35 |
Now taking upper class limits on x-axis and their respective cumulative frequency on y-axis we may draw its ogive as following
Now mark the point A whose ordinate is 17.5 its x-coordinate is 46.5. So median of this data is 46.5.
We may observe that difference between two consecutive upper class limits is 2. Now we may obtain class marks with their respective frequencies as below
| Weight (in kg) |
Frequency (f) | Cumulative frequency |
| Less than 38 | 0 | 0 |
| 38 – 40 | 3 – 0 = 3 | 3 |
| 40 – 42 | 5 – 3 = 2 | 5 |
| 42 – 44 | 9 – 5 = 4 | 9 |
| 44 – 46 | 14 – 9 = 5 | 14 |
| 46 – 48 | 28 – 14 = 14 | 28 |
| 48 – 50 | 32 – 28 = 4 | 32 |
| 50 – 52 | 35 – 32 = 3 | 35 |
| Total (n) | 35 |
Now the cumulative frequency just greater than 
is 28 belonging to class interval 46 – 48
Median class = 46 – 48
Lower class limit (l) of median class = 46
Frequency (f) of median class = 14
Cumulative frequency (cf) of class preceding median class = 14
Class size (h) = 2
So median of this data is 46.5
Hence, value of median is verified.
Question 3.
The following table gives production yield per hectare of wheat of 100 farms of a village.
| Production yield (in kg/ha) | 50 – 55 | 55 – 60 | 60 – 65 | 65 – 70 | 70 – 75 | 75 – 80 |
| Number of farms | 2 | 8 | 12 | 24 | 38 | 16 |
Change the distribution to a more than type distribution and draw ogive.
We can obtain cumulative frequency distribution of more than type as following –
| Production yield (lower class limits) |
Cumulative frequency |
| more than or equal to 50 | 100 |
| more than or equal to 55 | 100 – 2 = 98 |
| more than or equal to 60 | 98 – 8 = 90 |
| more than or equal to 65 | 90 – 12 = 78 |
| more than or equal to 70 | 78 – 24 = 54 |
| more than or equal to 75 | 54 – 38 = 16 |
Now taking lower class limits on x-axis and their respective cumulative frequencies on y-axis we can obtain its ogive as following.
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